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我想每毫秒调用一次 pollkey 函数并每秒增加一次时间变量（timeloc）。我认为如果我call pollkey在延迟子例程中添加一个它应该可以工作，为什么它不工作？

        .equ    delaycount,     16911 #set right delay value here!
        .text                   # Instructions follow
        .global delay           # Makes "main" globally known

delay:  beq     r4,r0,fin       # exit outer loop
        movi    r8,delaycount   # delay estimation for 1ms

inner:  beq     r8,r0,outer     # exit from inner loop
        subi    r8,r8,1         # decrement inner counter
        br      inner

outer:  subi    r4,r4,1         # decrement outer counter
        call    pollkey
        br      delay


fin:    ret

上面我用C运行的是

#include <stdio.h>
#include "system.h"
#include "altera_avalon_pio_regs.h"
extern void puttime(int* timeloc);
extern void puthex(int time);
extern void tick(int* timeloc);
extern void delay(int millisec);
extern int hexasc(int invalue);

#define TRUE 1
#define KEYS4 ( (unsigned int *) 0x840 )

int timeloc = 0x5957; /* startvalue given in hexadecimal/BCD-code */
int RUN = 1;

void pollkey() {
    int action = IORD_ALTERA_AVALON_PIO_DATA(DE2_PIO_KEYS4_BASE);
    if (action == 7) {
        timeloc = 0x0;
    } else if (action == 13) {
        RUN = 0;
    } else if (action == 14) {
        RUN = 1;
    } else if (action == 11) {
        tick(&timeloc);
    }
}

int main() {
    while (TRUE) {
        puttime(&timeloc);
        delay(1);
        IOWR_ALTERA_AVALON_PIO_DATA(DE2_PIO_REDLED18_BASE, timeloc);
        if (RUN == 1) {
            tick(&timeloc);
            puthex(timeloc);
        }

    }
    return 0;
}

int hex7seg(int digit) {
    int trantab[] = { 0x40, 0x79, 0x24, 0x30, 0x19, 0x12, 0x02, 0x78, 0x00,
            0x10, 0x08, 0x03, 0x46, 0x21, 0x06, 0x0e };
    register int tmp = digit & 0xf;
    return (trantab[tmp]);
}

void puthex(int inval) {
    unsigned int hexresult;
    hexresult = hex7seg(inval);
    hexresult = hexresult | (hex7seg(inval >> 4) << 7);
    hexresult = hexresult | (hex7seg(inval >> 8) << 14);
    hexresult = hexresult | (hex7seg(inval >> 12) << 21);
    IOWR_ALTERA_AVALON_PIO_DATA(DE2_PIO_HEX_LOW28_BASE, hexresult);
}

int hex7seg2(int digit) {
    int trantab[] = { 0x40, 0x79, 0x24, 0x30, 0x19, 0x12, 0x02, 0x78, 0x00,
            0x10, 0x08, 0x03, 0x46, 0x21, 0x06, 0x0e };
    register int tmp = digit & 0xf0;
    return (trantab[tmp]);
}

如果它每秒轮询一次，但我想每毫秒轮询一次，我认为我能做到的唯一方法是从 dely 子例程调用 pollkey，但如果我这样做，它就像什么也没发生。你能帮助我吗？我之前问过怎么做，只有在我认为答案应该是更改程序集时才在 C 中得到答案。

如何开发这个算法？

更新

br使用而不是我得到了更好的结果，call所以我必须检查差异。这就是我正在使用的效果更好的方法：

        .equ    delaycount,     16911 #set right delay value here!
        .text                   # Instructions follow
        .global delay           # Makes "delay" globally known

delay:  beq     r4,r0,fin       # exit outer loop
        movi    r8,delaycount   # delay estimation for 1ms

inner:  beq     r8,r0,outer     # exit from inner loop
        subi    r8,r8,1         # decrement inner counter
        br      inner

outer:  subi    r4,r4,1         # decrement outer counter
        br      pollkey
        br      delay


fin:    ret

现在唯一的问题是时钟走得太快了。

更新

我想我通过引入一个计数器来解决它，该计数器会在几秒钟内打破模 1000：

#include <stdio.h>
#include "system.h"
#include "altera_avalon_pio_regs.h"
extern void puttime(int* timeloc);
extern void puthex(int time);
extern void tick(int* timeloc);
extern void delay(int millisec);
extern int hexasc(int invalue);

#define TRUE 1
#define KEYS4 ( (unsigned int *) 0x840 )

int timeloc = 0x5957; /* startvalue given in hexadecimal/BCD-code */
int RUN = 0;

void pollkey() {
    int action = IORD_ALTERA_AVALON_PIO_DATA(DE2_PIO_KEYS4_BASE);
    if (action == 7) {
        timeloc = 0x0;
        puttime(&timeloc);
        puthex(timeloc);
    } else if (action == 13) {
        RUN = 0;
    } else if (action == 14) {
        RUN = 1;
    } else if (action == 11) {
        tick(&timeloc);
        puttime(&timeloc);
        puthex(timeloc);
        delay(1000);
    }
}

int main() {
    int counter = 0;
    while (TRUE) {
        pollkey();
        delay(1);
        ++counter;
        if (counter % 1000 == 0) {
            IOWR_ALTERA_AVALON_PIO_DATA(DE2_PIO_REDLED18_BASE, timeloc);
            if (RUN == 1) {
                tick(&timeloc);
                puttime(&timeloc);
                puthex(timeloc);
            }
        }

    }
    return 0;
}

int hex7seg(int digit) {
    int trantab[] = { 0x40, 0x79, 0x24, 0x30, 0x19, 0x12, 0x02, 0x78, 0x00,
            0x10, 0x08, 0x03, 0x46, 0x21, 0x06, 0x0e };
    register int tmp = digit & 0xf;
    return (trantab[tmp]);
}

void puthex(int inval) {
    unsigned int hexresult;
    hexresult = hex7seg(inval);
    hexresult = hexresult | (hex7seg(inval >> 4) << 7);
    hexresult = hexresult | (hex7seg(inval >> 8) << 14);
    hexresult = hexresult | (hex7seg(inval >> 12) << 21);
    IOWR_ALTERA_AVALON_PIO_DATA(DE2_PIO_HEX_LOW28_BASE, hexresult);
}

int hex7seg2(int digit) {
    int trantab[] = { 0x40, 0x79, 0x24, 0x30, 0x19, 0x12, 0x02, 0x78, 0x00,
            0x10, 0x08, 0x03, 0x46, 0x21, 0x06, 0x0e };
    register int tmp = digit & 0xf0;
    return (trantab[tmp]);
}


        .equ    delaycount,     16911 #set right delay value here!
        .text                   # Instructions follow
        .global delay           # Makes "delay" globally known

delay:  beq     r4,r0,fin       # exit outer loop
        movi    r8,delaycount   # delay estimation for 1ms

inner:  beq     r8,r0,outer     # exit from inner loop
        subi    r8,r8,1         # decrement inner counter
        br      inner

outer:  subi    r4,r4,1         # decrement outer counter
        br      delay


fin:    ret
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1 回答 1

1

在特定时间尺度上在嵌入式系统中运行例程的正确方法是使用硬件定时器并将操作（假设它相对较小！）放入该定时器的中断服务例程中，或者使用该函数来告诉更长的例程运行在后台循环中。

固定delay()功能将导致

  • 可变延迟 - 延迟函数已经是使用指令周期的近似值，如果你跑到中间调用另一个函数，有时它只会变得更糟
  • 不准确的 1ms 轮询 - 不准确的程度可能对您的目的来说不够好，也可能不够好。

一旦你有了一个 1ms 的计时器，你就可以从中运行pollkey函数，也可以准确地安排更长的事件。

当然，那时你已经开始构建一个 RTOS，所以你可能还是想使用一个 :)

于 2013-09-07T18:38:51.237 回答